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Limit of the function 9-x

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The solution

You have entered [src] 
 lim (9 - x)
x->oo
lim⁡x→∞(9−x)\lim_{x \to \infty}\left(9 - x\right) 
Limit(9 - x, x, oo, dir='-')
Detail solution
Let's take the limit
lim⁡x→∞(9−x)\lim_{x \to \infty}\left(9 - x\right)
Let's divide numerator and denominator by x:
lim⁡x→∞(9−x)\lim_{x \to \infty}\left(9 - x\right) =
lim⁡x→∞(−1+9x1x)\lim_{x \to \infty}\left(\frac{-1 + \frac{9}{x}}{\frac{1}{x}}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
lim⁡x→∞(−1+9x1x)=lim⁡u→0+(9u−1u)\lim_{x \to \infty}\left(\frac{-1 + \frac{9}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{9 u - 1}{u}\right)
=
−1+0⋅90=−∞\frac{-1 + 0 \cdot 9}{0} = -\infty

The final answer:
lim⁡x→∞(9−x)=−∞\lim_{x \to \infty}\left(9 - x\right) = -\infty
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
02468-8-6-4-2-1010-2020
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Rapid solution [src] 
-oo
−∞-\infty
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(9−x)=−∞\lim_{x \to \infty}\left(9 - x\right) = -\infty
lim⁡x→0−(9−x)=9\lim_{x \to 0^-}\left(9 - x\right) = 9
More at x→0 from the left
lim⁡x→0+(9−x)=9\lim_{x \to 0^+}\left(9 - x\right) = 9
More at x→0 from the right
lim⁡x→1−(9−x)=8\lim_{x \to 1^-}\left(9 - x\right) = 8
More at x→1 from the left
lim⁡x→1+(9−x)=8\lim_{x \to 1^+}\left(9 - x\right) = 8
More at x→1 from the right
lim⁡x→−∞(9−x)=∞\lim_{x \to -\infty}\left(9 - x\right) = \infty
More at x→-oo
The graph
Limit of the function 9-x
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